Problem: Alice has a cute cat. To keep her cat fit, Alice wants to design an exercising walk for her cat! 

Initially, Alice's cat is located in a cell (x,y) of an infinite grid. According to Alice's theory, cat needs to move: 

  * exactly a steps left: from (u,v) to (u-1,v); 
  * exactly b steps right: from (u,v) to (u+1,v); 
  * exactly c steps down: from (u,v) to (u,v-1); 
  * exactly d steps up: from (u,v) to (u,v+1). 



Note that the moves can be performed in an arbitrary order. For example, if the cat has to move 1 step left, 3 steps right and 2 steps down, then the walk right, down, left, right, right, down is valid.

Alice, however, is worrying that her cat might get lost if it moves far away from her. So she hopes that her cat is always in the area [x_1,x_2]× [y_1,y_2], i.e. for every cat's position (u,v) of a walk x_1 ≤ u ≤ x_2 and y_1 ≤ v ≤ y_2 holds.

Also, note that the cat can visit the same cell multiple times.

Can you help Alice find out if there exists a walk satisfying her wishes?

Formally, the walk should contain exactly a+b+c+d unit moves (a to the left, b to the right, c to the down, d to the up). Alice can do the moves in any order. Her current position (u, v) should always satisfy the constraints: x_1 ≤ u ≤ x_2, y_1 ≤ v ≤ y_2. The staring point is (x, y).

You are required to answer t test cases independently.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^3) — the number of testcases. 

The first line of each test case contains four integers a, b, c, d (0 ≤ a,b,c,d ≤ 10^8, a+b+c+d ≥ 1).

The second line of the test case contains six integers x, y, x_1, y_1, x_2, y_2 (-10^8 ≤ x_1≤ x ≤ x_2 ≤ 10^8, -10^8 ≤ y_1 ≤ y ≤ y_2 ≤ 10^8).

Output

For each test case, output "YES" in a separate line, if there exists a walk satisfying her wishes. Otherwise, output "NO" in a separate line. 

You can print each letter in any case (upper or lower).

Example

Input


6
3 2 2 2
0 0 -2 -2 2 2
3 1 4 1
0 0 -1 -1 1 1
1 1 1 1
1 1 1 1 1 1
0 0 0 1
0 0 0 0 0 1
5 1 1 1
0 0 -100 -100 0 100
1 1 5 1
0 0 -100 -100 100 0


Output


Yes
No
No
Yes
Yes
Yes

Note

In the first test case, one valid exercising walk is $$$(0,0)→ (-1,0) → (-2,0)→ (-2,1) → (-2,2)→ (-1,2)→(0,2)→ (0,1)→ (0,0) → (-1,0)$$$

Solution (in JAVA):

Solution:


import java.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;

public class Main {
	static final int INF = (int) (1e9 + 10);
	static final int MOD = (int) (1e9 + 7);
	// static final int N = (int) (4e5 + 5);
	// static ArrayList<Integer>[] graph;
	// static boolean visited[];
	// static int size[];

	public static void main(String[] args) throws NumberFormatException, IOException {
		FastReader sc = new FastReader();
		PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		// Scanner scn = new Scanner(System.in);
		//
		int t = sc.nextInt();
		while (t-- > 0) {
			long a = sc.nextLong();
			long b = sc.nextLong();
			long c = sc.nextLong();
			long d = sc.nextLong();
			long x = sc.nextLong();
			long y = sc.nextLong();
			long x1 = sc.nextLong();
			long y1 = sc.nextLong();
			long x2 = sc.nextLong();
			long y2 = sc.nextLong();
			long fx = (b - a)+x;
			long fy = (d - c)+y;
		//	 System.out.println(fx+" "+fy+" "+x1+" "+x2+" "+y1+" "+y2);
			if ((x1 - x2) == 0 && (a > 0 || b > 0)) {
				pr.println("No");
			} else if ((y1 - y2) == 0 && (c > 0 || d > 0)) {
				pr.println("No");
			} else if (fx >= x1 && fx <= x2 && fy >= y1 && fy <= y2) {
				pr.println("Yes");
			} else
				pr.println("No");
		}

		//
		// coded to perfection by Rohan Mukhija
		pr.flush();
		pr.close();
	}

	/*
	 * Template From Here
	 */
	private static boolean possible(long[] arr, int[] f, long mid, long k) {
		long c = 0;

		for (int i = 0; i < arr.length; i++) {
			if ((arr[i] * f[f.length - i - 1]) > mid) {
				c += (arr[i] - (mid / f[f.length - i - 1]));
			}
		}
		// System.out.println(mid+" "+c);
		if (c <= k)
			return true;
		return false;
	}

	public static int lowerBound(ArrayList<Integer> array, int length, long value) {
		int low = 0;
		int high = length;
		while (low < high) {
			final int mid = (low + high) / 2;
			if (value <= array.get(mid)) {
				high = mid;
			} else {
				low = mid + 1;
			}
		}
		return low;
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

	static long gcd(long a, long b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}

	static long lcm(long a, long b) {
		return (a * b) / gcd(a, b);
	}

	public static void sortbyColumn(int[][] att, int col) {
		// Using built-in sort function Arrays.sort
		Arrays.sort(att, new Comparator<int[]>() {

			@Override
			// Compare values according to columns
			public int compare(final int[] entry1, final int[] entry2) {

				// To sort in descending order revert
				// the '>' Operator
				// if (entry1[col] > entry2[col])
				// return 1;
				// else //(entry1[col] >= entry2[col])
				// return -1;
				return new Integer(entry1[col]).compareTo(entry2[col]);
				// return entry1[col].
			}
		}); // End of function call sort().
	}

	static class pair {
		int V, I;

		pair(int v, int i) {
			V = v;
			I = i;
		}
	}

	public static int[] swap(int data[], int left, int right) {
		int temp = data[left];
		data[left] = data[right];
		data[right] = temp;
		return data;
	}

	public static int[] reverse(int data[], int left, int right) {
		while (left < right) {
			int temp = data[left];
			data[left++] = data[right];
			data[right--] = temp;
		}
		return data;
	}

	public static boolean findNextPermutation(int data[]) {
		if (data.length <= 1)
			return false;

		int last = data.length - 2;
		while (last >= 0) {
			if (data[last] < data[last + 1]) {
				break;
			}
			last--;
		}
		if (last < 0)
			return false;

		int nextGreater = data.length - 1;
		for (int i = data.length - 1; i > last; i--) {
			if (data[i] > data[last]) {
				nextGreater = i;
				break;
			}
		}
		data = swap(data, nextGreater, last);
		data = reverse(data, last + 1, data.length - 1);
		return true;
	}

	static long nCr(long n, long r) {
		if (n == r)
			return 1;
		return fact(n) / (fact(r) * fact(n - r));
	}

	static long fact(long n) {
		long res = 1;
		for (long i = 2; i <= n; i++)
			res = res * i;
		return res;
	}

	/*
	 * static boolean prime[] = new boolean[1000007];
	 * 
	 * public static void sieveOfEratosthenes(int n) {
	 * 
	 * for (int i = 0; i < n; i++) prime[i] = true; * for (int p = 2; p * p <=
	 * n; p++) { // If prime[p] is not changed, then it is a prime if (prime[p]
	 * == true) { // Update all multiples of p for (int i = p * p; i <= n; i +=
	 * p) prime[i] = false; } }
	 * 
	 * // Print all prime numbers // for(int i = 2; i <= n; i++) // { //
	 * if(prime[i] == true) // System.out.print(i + " "); // } }
	 */

	static long power(long fac2, int y, int p) {
		long res = 1;
		fac2 = fac2 % p;
		while (y > 0) {
			if (y % 2 == 1)
				res = (res * fac2) % p;
			fac2 = (fac2 * fac2) % p;
		}
		return res;
	}

	static long modInverse(long fac2, int p) {
		return power(fac2, p - 2, p);
	}

	static boolean isPrime(int n) {
		// Corner cases
		if (n <= 1)
			return false;
		if (n <= 3)
			return true;

		// This is checked so that we can skip
		// middle five numbers in below loop
		if (n % 2 == 0 || n % 3 == 0)
			return false;

		for (int i = 5; i * i <= n; i = i + 6)
			if (n % i == 0 || n % (i + 2) == 0)
				return false;

		return true;
	}

	public static BigInteger lcmm(String a, String b) {
		BigInteger s = new BigInteger(a);
		BigInteger s1 = new BigInteger(b);
		BigInteger mul = s.multiply(s1);
		BigInteger gcd = s.gcd(s1);
		BigInteger lcm = mul.divide(gcd);
		return lcm;
	}

}